SOLUTION: Especially for the type problem listed below I get messed up on the part where I have to compensate for the distributive law, pretty much getting the common factor that gets thrown

Algebra ->  Equations -> SOLUTION: Especially for the type problem listed below I get messed up on the part where I have to compensate for the distributive law, pretty much getting the common factor that gets thrown      Log On


   

Question 1001575: Especially for the type problem listed below I get messed up on the part where I have to compensate for the distributive law, pretty much getting the common factor that gets thrown back in is difficult for me. Is there any way to easily know what should go where. It's hard for me to see exactly how it should be plugged back. Are there any tips/tricks for this to see it?

Here it the problem:
f(x) = (5/3)x^(2/3) + (10/3)x^(-1/3)
the answer is:
(5/3)x^(1/3) [x + 2] = 0
Please be detailed I really need to understand how this works. I can do other distribute problems well, but ones like these are harder for me.
Thank you
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+%285%2F3%29x%5E%282%2F3%29+%2B+%2810%2F3%29x%5E%28-1%2F3%29...........first factor out +%285%2F3%29

f%28x%29+=+%285%2F3%29%28x%5E%282%2F3%29+%2B2x%5E%28-1%2F3%29%29.....recall that x%5E%28-1%2F3%29=1%2Fx%5E%281%2F3%29

f%28x%29+=+%285%2F3%29%28x%5E%282%2F3%29+%2B2%2Fx%5E%281%2F3%29%29.....common denominator is x%5E%281%2F3%29



f%28x%29+=+%285%2F3%29%28x%5E%283%2F3%29%2Fx%5E%281%2F3%29+%2B2%2Fx%5E%281%2F3%29%29

f%28x%29+=+%285%2F3%29%28x%2Fx%5E%281%2F3%29+%2B2%2Fx%5E%281%2F3%29%29

f%28x%29+=+%285%2F3%29%28%28x+%2B2%29%2Fx%5E%281%2F3%29%29

f%28x%29+=+%285%2F3%29x%5E%28-1%2F3%29%28x+%2B2%29



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Especially for the type problem listed below I get messed up on the part where I have to compensate for the distributive law, pretty much getting the common factor that gets thrown back in is difficult for me. Is there any way to easily know what should go where. It's hard for me to see exactly how it should be plugged back. Are there any tips/tricks for this to see it?

Here it the problem:
f(x) = (5/3)x^(2/3) + (10/3)x^(-1/3)
the answer is:
(5/3)x^(1/3) [x + 2] = 0
Please be detailed I really need to understand how this works. I can do other distribute problems well, but ones like these are harder for me.
Thank you


If you look at the expression, you can see that 5%2F3 is common to each polynomial, and so is: highlight%28highlight%28x%5E%28-1%2F3%29%29%29

Therefore, the GCF of the expression is: highlight%28highlight%28highlight%28%285%2F3%29x%5E%28-+1%2F3%29%29%29%29

We now divide the 1st polynomial by the GCF: highlight%28highlight%28%285%2F3%29x%5E%282%2F3%29%29%29%22%F7%22highlight%28highlight%28%285%2F3%29x%5E%28-+1%2F3%29%29%29 = highlight%28highlight%28x%5E%282%2F3+-+-+1%2F3%29%29%29 = highlight%28highlight%28x%5E%283%2F3%29%29%29, or x



We now divide the 2nd polynomial by the GCF: highlight%28highlight%28%2810%2F3%29x%5E%28-+1%2F3%29%29%29%22%F7%22highlight%28highlight%28%285%2F3%29x%5E%28-+1%2F3%29%29%29 = highlight%28highlight%282+%2A+x%5E%28-+1%2F3+-+-+1%2F3%29%29%29 = 2+%2A+x%5E0 = 2 * 1, or + 2

We now have: highlight_green%28highlight_green%28%285%2F3%29x%5E%28-+1%2F3%29%28x+%2B+2%29%29%29



Your answer is incorrect!! The first factor has -+%281%2F3%29 as its exponent i/o 1%2F3